Pipes! and no, not the one's for smoking
Introduction The pipes I will be discussing in this article are not used in plumbing, nor do you smoke tobacco out of them, I will be talking about pipes that produce sound. These pipes are found in instruments such as an organ. Greek engineer Ktesibios first introduced the organ in 3rd century B.C. He wanted to be able to play more than one wind instrument at a time. Wind instruments have been around since even before the organ. There is also such a wide variety of wind instruments, ranging from the piccolo to the tuba. All of the wind instruments vary in shape and size, but the one thing all these instruments have in common is how their sound is produced. As you know, sound is generated by vibrations, so when Air is blown into one end of the pipe or tube and then bounces off of the sides, the air vibrates. When the air inside the tube vibrates at the same frequency, or in resonance, with the vibration of you lips, a sound is produced. You can achieve different sounds by changing the length or width of the instrument. You can ultimately create a sound that no one has ever heard before, but all you need is a little imagination. What Happens In Different Pipes 'Closed Pipe Resonator:' In this world there are two kinds of pipes, closed pipes, and open pipes. Closed pipes are open on one end and closed on the other. Open pipes allow substance to flow through both ends. In a closed pipe when a sound enters the pipe, it bounces off the bottom and moves back towards the top. A great example of a closed-pipe resonator is a tuning fork held over a hollow tube that is placed in water.The tuning fork creates a sound wave at a specific frequency. The hollow tube allows the sound to resinate with the air inside it, and finally the water creates a closed end for the hollow tube. When you hit the tuning fork, you hold it above the hollow tube and move the tube up and down in the water until you can hear a sound. What you will start to notice, is that as you raise and lower the hollow tube the sound will get louder or softer. This is because at certain heights the wave interferes with itself creating a standing wave. 'Open Pipe Resonator:' In an open pipe however, there is no end for the sound wave to bounce back from. This closed end is not needed because due to the pressure of the wave, it will reverberate back of the open end of the pipe. A great example of this is the trumpet again. It has two open ends and it still resonates. If it didn’t, you wouldn’t hear any sound coming out of it. Length Of Waves 'In Closed Pipes:' In a closed pipe, as I said earlier, one end is open and the other is closed. This means that the wave can move the most at the opening and the least at the closed portion of the pipe. What we can derive from this is that there is a node at the closed part of the pipe, and there is an antinode at the opening. If you begin to increase the frequency you will notice that more resonance lengths will appear in half wave length intervals. The first frequency (n=1) is known as the fundamental frequency. In a closed pipe the fundamental frequency is one-forth of a wavelength. The next value for n is called the first overtone. So at the first overtone (n=2) the frequency is going to be one-half wavelength longer than the fundamental frequency, or three-forth’s of a wave length. This pattern will go on depending on frequency values. This change in Wavelength can be represented by the formula: wavelength = 4L/(2n-1) 'In Open Pipes:' In an open pipe, both sides are open, so this means the maximum amount of movement can occur at both ends of the pipe. So this means the fundamental frequency for an open pipe is one-halve of a wavelength. Open pipes also have additional resonance lengths, all at one-half wavelength intervals. This change in wavelength can be represented by the formula: l=(n/2) wavelength or wavelength=2l/n where n=number of nodes l= length of pipe Sample Problems 1. If a bugle were straightend out, it would be 2.65m long. a. If the speed of sound is 343m/s find the lowest frequency that is resonant in a bugle b. Find the next two higher resonant frequencies in the bugle 2 This is a fairly simple problem. first for a. what you must do it remeber the equation f=v/wavelength. all you do after that is add the n=1 to the equation, and plug in know values, and that will give you f. Known: v=343m/s wavelenth=2L or 5.3m f=v/wavelength f=343/5.3 f=65.0 Hz for b. all you have to use is the values of n=2 and n=3 instead of just using n=1. For n=2: Known: v=343m/s wavelength= L or 2.65 f=v/wavelegnth f=342/2.65 f=129 Hz For n=3: Known: v=343m/s wavelength= 2L or 5.3m f=v/wavelength f=3v/2L f=1029/5.3 f=194.0 Hz References http://rockefeller.uchicago.edu/organ.htmlhttp://rockefeller.uchicago.edu/organ.html -information on organs 2Zitzewitz, Paul W. Physics Principles and Problems. New York: Glencoe McGraw-Hill, 1999. 357-361.-Text book full of information 3Closed -Wikia Page on closed pipes 4Open -Wikia page on open pipes